Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <stdio.h>

const int MAX_N = 3005;

typedef long double real;

int n;
real p[MAX_N],dp[MAX_N][MAX_N];

int main()
{
	scanf("%d",&n);
	for (int i=1; i<=n; i++)
		scanf("%Lf",p+i);
	dp[0][0]=1;
	for (int i=1; i<=n; i++)
	{
		dp[i][0]=(1-p[i])*dp[i-1][0];
		for (int h=1; h<=i; h++)
			dp[i][h]=p[i]*dp[i-1][h-1]+(1-p[i])*dp[i-1][h];
	}
	real ans=0;
	for (int h=n; h>n-h; h--)
		ans+=dp[n][h];
	printf("%.10Lf\n",ans);
	return 0;
}